# Two equal angles

Let $$\triangle ABC$$ be a triangle with circumcircle (O) P is on BC, PA is tangent to (O). $$E \in PO$$, $$D \in BE$$$$AD \bot AB$$. Prove that $$\angle EAB=\angle ACD$$. Please provide an elementary proof for this.

and

$$\angle CBF =\angle DCA \Leftrightarrow \frac{\sin{\angle BAE}}{\sin{\angle EAO}} =\frac{\sin{\angle ACD}}{\sin{\angle DCH}} =\frac{AD*HC}{DH*AC}$$ $$\frac{\sin{\angle BAE}}{\sin{\angle EAO}} =\frac{AD*CH*\sin{\angle EOB}}{DH*BA*\sin{\angle AOE}}$$ It only need to prove that: $$\frac{AB*HC}{BH*AC} =\frac{\sin{\angle EOB}}{\sin{\angle AOE}} =\frac{\sin{\angle OPB}*\sin{\angle PAB}}{\sin{\angle OPA}*\sin{\angle PBA}}$$ But I have no idea about it. Any ideas?

• Please show your effort, otherwise your question would be likely put on hold – Oldboy Feb 10 '19 at 6:53
• @qsa Math SE is not Brilliant. You can show your work or also show the inspiration to the question (more details here). – Toby Mak Feb 10 '19 at 8:21
• The sentence "I tried" is not sufficient. In your attempts, you have probably obtained some results beyond the text of the question, which you haven't yet been able to make use of. Do write them out. – Joce Feb 14 '19 at 12:47
• If you're still working on this: you have a property of angles formed by tangents to circumcircle at triangle vertices (see en.wikipedia.org/wiki/Circumscribed_circle#Angles ) – Joce Feb 25 '19 at 12:22